Question #67c37

1 Answer
Jul 19, 2017

Answer:

We needs approx. #15*g# of #NaNO_2(s)#, and almost #150*mL# of #"20%(w/w)"# #"ammonium chloride"#.

Explanation:

Your first step is to write a stoichiometric equation.....this is absolutely vital.

#NH_4NO_2(s) +Deltararr N_2(g)uarr + 2H_2O(l)#

This is a comproportionation reaction, in which nitrogen is simultaneously oxidized (from #N(-III)# in #NH_4^+#), and reduced (from #N(+III)# in #NO_2^(-)#) to give zerovalent dinitrogen gas.....

And given the clear stoichiometry, each equiv dinitrogen gas corresponds to one equiv of #"ammonium nitrite"#.

Now, given the Ideal gas equation, and we may reasonably assume ideality, then #n=(PV)/(RT)#, and we fill in the dots......

#"Moles of dinitrogen"-=((750*mm*Hg)/(760*mm*Hg*atm^-1)xx5.57*L)/(0.0821*(L*atm)/(K*mol)xx300*K)#

#=0.223*1/(mol^-1)=0.223/(1/(mol))=0.223*mol#

And thus there were about #0.223*mol# with respect to #"ammonium nitrite"# given the 1:1 molar stoichiometry which is explicitly specified by the decomposition reaction.

And thus there were #0.223*mol# of sodium nitrite, i.e. #0.223*molxx69.0*g*mol^-1=15.4*g#.

And we use these data to tell us that #"20%(w/w)"# #"ammonium chloride"#. has a concentration of #83.1*g*L^-1# with respect to #NH_4Cl#, i.e. #1.55*mol*L^-1#.

And so we needs #(0.223*mol)/(1.55*mol*L^-1)xx1000*mL*L^-1#

#=143.9*mL#.

Are you with me? I urge you to go thru the equations yourself, and reproduce it (and you might find an error on my part). Part of the problem is that your question assumed knowledge of data that were not quoted in the question. (These data should have been quoted!).